The time taken for the completion of 90% of a first-order reaction is ‘t’ min. What is the time (in seconds) taken for the completion of 99% of the reaction?

120 t

The correct answer is option 3. 120 t.

Case I:

Given, the time taken for the completion of 90% of a first-order reaction is ‘t’ min

As the reaction is 90% complete, \(a = 100, \text{ and }a − x = 100 −90 = 10\)

So,

\(t_{90} = \frac{2.303}{k}log\frac{100}{10}\)

or, \(t = \frac{2.303}{k}log(10)\)     [since t₉₀ = t min]

or, \(t = \frac{2.303}{k} × 1\) ------(i)

Case II:

When the reaction is 99% complete, then

 \(a = 100, \text{ and }a − x = 100 −  99 = 1\)

So,

\(t_{99} = \frac{2.303}{k}log\frac{100}{1}\)

or, \(t_{99} = \frac{2.303}{k}log(100)\)

or, \(t_{99} = \frac{2.303}{k} × 2\) ------(ii)

Dividing equation (ii) by (i), we get

\(\frac{t_{99}}{t} = \frac{2}{1}\)

or, \(\frac{t_{99}}{t} =2min\)

or, \(t_{99} = 2t\text{ min}\)

\(∴ t_{99} = 120t \text{ sec}\)     [Since, 1 min = 60 sec]